Go Premium for a chance to win a PS4. Enter to Win

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 1439
  • Last Modified:

Backwards Euler method

The final part of a question presents the "Backwards Euler method", as:

x(n+1) = x(n) + h * f( (n+1) * h,  x(n+1) )              (where f(t,x) = dx/dt)

It then says to apply this method to the linear equation dx/dt=x, and show that the method converges to the true solution x(t)=e^t as t->infinity.

It is obvious that the term (1+t/n)^n will turn up at some stage (seeing as the limit of it, as t->infinity, is e^t). But I'm struggling to get started. My problem is that in order to find the derivative at x(n+1), I must find what x(n+1) is; but to find x(n+1), I need to know the derivative at x(n+1)...

So, can someone at least get me past this hurdle, then I will hopefully be able to figure the rest for myself.

An earlier part of this question might be relevent; it says to apply Euler's method to dx/dt=kx. For which I get:

x(n)=x0(1+hk)^n

...

Thanks
0
Beta07
Asked:
Beta07
  • 4
1 Solution
 
ozoCommented:
isn't dx/dt=x the same as  dx/dt=kx with k=1
0
 
Beta07Author Commented:
Indeed. I suppose you're suggesting that I use:

x(n)=x0(1+hk)^n

(with k=1), to find x(n+1)? I thought about this, but (although it's probably the right way to go), it doesn't really make much sense to me:
Surely by incorporating the standard Euler method into the backwards one, you'd lose some of the advantages of using the backwards method in the first place?

Either way, I'll assume it's the correct method. Which gives me:

x(1) = x(0) + h*x(0)(1+h)
x(2) = x(1) + h*x(0)(1+h)²
       = x(0) + h*x(0)(1+h) + h*x(0)(1+h)²
x(3) = x(2) + h*x(0)(1+h)³
       = x(0) + h*x(0)(1+h) + h*x(0)(1+h)^2 + h*x(0)(1+h)³
...
x(n+1) = x(0){1+h(1+h) + h(1+h)² + h(1+h)³ + ... + h(1+h)^n}
           = x(0){(1+h)^(n+1) - h)}

(Using the equation for the partial sum of a geometric series (and rearranging), in that last step).

Now, this looks quite promising:

x(n+1) = x(0){(1+h)^(n+1) - h)}
=>
x(n) = x(0){(1+h)^n - h}

The question says to use the substitution: h=t/n; thus

x(n) = x(0){(1+t/n)^n - h}

So we're almost there:  the limit as h->0 (or n->infinity)* of
{(1+t/n)^n - h}
is:
e^t

However, I'm not sure how to properly finish this question off. I don't think it would make much sense to say:

lim[h->0]  x(n)/x(0)  =  e^t

So, I guess I'm looking for a rearrangement of:

x(n) = x(0){(1+t/n)^n - h}

such that when I take the limit of both sides, I get x(t) on one side, and e^t on the other....?


Thank you

* In my original question (and also on the question sheet), it says "as t->infinity (equivalently, h->0)"; I can only assume they mean "n->infinity,.. h->0".
0
 
Beta07Author Commented:
(Please ignore the  characters)
0
 
Beta07Author Commented:
>>
So, I guess I'm looking for a rearrangement of:

x(n) = x(0){(1+t/n)^n - h}

such that when I take the limit of both sides, I get x(t) on one side, and e^t on the other....?
>>

Or, I suspect, a rearrangement such that when I take the limit of both sides, I get dx/dt on one side, and e^t on the other..?
0
 
Beta07Author Commented:
Sorry,

> x(n+1) = x(0){1+h(1+h) + h(1+h)² + h(1+h)³ + ... + h(1+h)^n}
              = x(0){(1+h)^(n+1) - h)}

Should be

x(n) = x(0){1+h(1+h) + h(1+h)² + h(1+h)³ + ... + h(1+h)^n}
       = x(0){(1+h)^(n+1) - h)}

Thus

>>
x(n+1) = x(0){(1+h)^(n+1) - h)}
=>
x(n) = x(0){(1+h)^n - h}
>>

Is of course wrong.

So we've got:

x(n) = x(0){(1+h)^(n+1) - h)}
0

Featured Post

Hire Technology Freelancers with Gigs

Work with freelancers specializing in everything from database administration to programming, who have proven themselves as experts in their field. Hire the best, collaborate easily, pay securely, and get projects done right.

  • 4
Tackle projects and never again get stuck behind a technical roadblock.
Join Now